Question

Elevation in boiling point was 0.52^c when 6 g of a compound X was dissolved in 100g of water. Molecular mass of compound X is...... (Kb for water is 0.52 per 1000g of water).

Answer 1

Answer : The molecular weight of compound x is 8.76 g/mole.

Solution : Given,

$k_b=0.52Kkg/mole$    (for water)

Mass of solute = 6 g

Volume of solvent = 100 g

elevation constant = $0.52^oC=273+0.52=273.52K$

The formula used for depression in freezing point is,

$\Delta T_f=k_f\times m\\\Delta T_f=k_f\times \frac{w_{solute}}{M_{solute}\times w_{solvent}}$

where,

$\Delta T_f$ = elevation constant

m = molality

$k_b$ = molal elevation constant

Now put all the given values in this formula, we get

$273.52K=0.52Kkg/mole\times \frac{6g\times 1000}{M_{solute}\times 100kg}$

${M_{solute}=8.76g/mole$

Therefore, the molecular weight of compound x is 8.76 g/mole.

Answer 2

Answer:

∆Tb = kbm

mass of solute / molar mass of solvent ( g) × 1000

0.52 = 0.52 × 6 / m × 100 × 1000

60 g / mol